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devlian [24]
3 years ago
13

I really need Nitro type gold. So can anyone get it for me? My user name is Foxymadn117_YT. If you have a nitro type account i w

ill give you 2 million dollars for it. And thank you to anyone that can help me.
Mathematics
1 answer:
Darina [25.2K]3 years ago
7 0

Answer:

no

Step-by-step explanation:

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Tamiku [17]
1 divided by 10 is equal to 0.1 so you multiply 52 by 0.1 and you get an answer of 5.2 inches

6 0
4 years ago
(7−4n)⋅6 Apply the distributive property to create an equivalent expression.
avanturin [10]

Answer:

42-24n

Step-by-step explanation:

(7 - 4n) . 6

= (7)(6) - (4n)(6) ...........using distributive property, multiply each term by (6)

= 42 -24n (answer)

7 0
3 years ago
The lower base of the frustum shown is a square 8 ft on each side. The upper base is a square 4 ft on each side. If the altitude
Lemur [1.5K]
The Slant height is 7.21
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4 years ago
Nobody Knows the Answer HELP !!!!! Pleaseeeeeeeeee   CLICK & Help Me Solve !!!!!!!!!
sukhopar [10]
There are 10 total marbles and three are pink. So you are to determine the probability of choosing a pink marble, placing it back in the bag and choosing another pink marble.
The probability of choosing a pink marble is # of pink marbles to # of all marbles which is 3/10. Since we place the marble back in the bag, the probability is the same for the second time.
To determine the probability of both occurrences, we need to multiply 3/10 x 3/10 which is 9/100. This can be written as 9:100 as a ratio.
4 0
3 years ago
A model for the population in a small community after t years is given by P(t)=P0e^kt.
LUCKY_DIMON [66]
\bf \textit{Amount of Population Growth}\\\\
A=Ie^{rt}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
I=\textit{initial amount}\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\\
\end{cases}

a)

so, if the population doubled in 5 years, that means t = 5.  So say, if we use an amount for "i" or P in your case, to be 1, then after 5 years it'd be 2, and thus i = 1 and A = 2, let's find "r" or "k" in your equation.

\bf \textit{Amount of Population Growth}\\\\
A=Ie^{rt}\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &2\\
I=\textit{initial amount}\to &1\\
r=rate\\
t=\textit{elapsed time}\to &5\\
\end{cases}
\\\\\\
2=1\cdot e^{5r}\implies 2=e^{5r}\implies ln(2)=ln(e^{5r})\implies ln(2)=5r
\\\\\\
\boxed{\cfrac{ln(2)}{5}=r}\qquad therefore\qquad \boxed{A=e^{\frac{ln(2)}{5}\cdot t}} \\\\\\
\textit{how long to tripling?}\quad 
\begin{cases}
A=3\\
I=1
\end{cases}\implies 3=1\cdot e^{\frac{ln(2)}{5}\cdot t}

\bf 3=e^{\frac{ln(2)}{5}\cdot t}\implies ln(3)=ln\left( e^{\frac{ln(2)}{5}\cdot t} \right)\implies ln(3)=\cfrac{ln(2)}{5} t
\\\\\\
\cfrac{5ln(3)}{ln(2)}=t\implies 7.9\approx t

b)

A = 10,000, t = 3

\bf \begin{cases}
A=10000\\
t=3
\end{cases}\implies 10000=Ie^{\frac{ln(2)}{5}\cdot 3}\implies \cfrac{10000}{e^{\frac{3ln(2)}{5}}}=I
\\\\\\
6597.53955 \approx I
3 0
4 years ago
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